how is i^x=2 possible?
ฝัง
- เผยแพร่เมื่อ 8 มิ.ย. 2024
- Learn more complex numbers from Brilliant: 👉brilliant.org/blackpenredpen/ (20% off with this link!)
Can the power of the imaginary unit i ever give us 2? We know i^1=1, i^2=-1, i^3=-i, i^4=1, and so on. So is it possible for us to find x so that i^x=2? Well, the answer is yes but it is not real. #ComplexNumbers #Exponentiation #ImaginaryNumbers #Mathematics #blackpenredpen
0:00 We know i^1=1, i^2=-1, i^3=-i, i^4=1, and so on. So is it possible for us to have i^x=2?
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pls Can 2^x=0 in the field of complex num ?
@@ayoubbenchetioui6481 x is negative infinity
Please solve this equation: (-2)^x=2
Thank you
Where can I apply this useless knowledge?
We need more τ. e^-(τ/4 + nτ).
I really love it when you say, " I don't like to be on the bottom, I like to be on the top"🤗🤗🤗🤗
🤨
A very normal, totally not suspicious, comment
🤨📸
@@thexavier666 yea absolutely, no complications there right? 🤨
@@noreoli true, no complications 👌👌👌
Complex world is crazy.
I wil be find new world
Name is fantastic number
By saying crazy, you are underestimating the craziness of complex numbers.
Gerçekten bu inanılmaz. Kuantum mekaniğinde de çok önemlidir.🙂
I am not even good in real numbers
Because the world is complex by itself
"Please don't say 90 degrees, we are all adults here"
"Complex number is a pathway to many abilities some consider to be unnatural." - Chancellor Palpatine said during his complex analysis lecture
Of course!
In fact, a^b can always = c, for all Complex numbers {a,b,c} where {a,b,c} not = 0 or 1.
I’m scared to look up that proof
@@DroughtBee The proof is left as an exercise for the reader
@@maximilianarold I think there isn't really a proof. I think it's just an assumption. Like 5^3 will always be equal to a value. Lets call this value c. so 5^3 = c. As we know, c = 125, so 5^3 is 125. We knew what a and b were, but we didn't know what c was, but there was an answer. So what if we know the value of a and c, but not b. so i^b = 2. Last time we didn't know a varible, there was an answer, so there must be an answer again. That is my assumption on how he got it, but he might actually have some reason behind it...
Let a^b=exp(b*ln(a)), since exp: C -> C* is surjective, there is a complex number z such that exp(z)=c.
@@TheEGod. The proof is actually really easy. Suppose we want to find b such that a^b=c with a,c not equal to 0 or 1. Take a natural log of both sides to get b*ln(a)=ln(c). So b=ln(c)/ln(a) works. That's it.
Note that this is the same answer from this video. Here c=2 and a=i, and ln(i)=i*pi/2, so b=-2*i*ln(2)/pi.
The other case is to suppose we want find a such that a^b=c with b,c not equal to 0 or 1. This one is even easier: a = c^-b.
This one is fun. I actually found your e^(e^x) = 1 video first.
I took a different approach to solving this. I tried computing the log-base i of 2, which using log quotient rule, is equivalent to ln(2) / ln(i), the bottom part of which I was able to solve by getting the angle (or family of angles) that has a cosine of 0 and a sine of 1, which is pi/2. Thanks!
Awesome, thanks!!
@@blackpenredpenhello
I have been watching your videos for about 2 years now, now that Im in UNI, and im learning more and more about math, I can follow the videos much better, and I just love to see the progress, and the interesting things you show here on youtube
I'd add a simplification. You can pull the 2 inside the ln as an exponent, so 2*ln(2) = ln(2^2) = ln(4). It makes the result a little prettier :D
Make π τ again.
Very nice explanation about why the +4n part is needed to complete the answer. i raised to any multiple of 4 will always result in 1.
please upload more, I really enjoy your videos
Great video... much appreciated. Your info shared and your style... and your nice manner.
Hey, nice video, I've got a fun challenge for you: Determine all positive integer pairs (p, q) for that p^q + q^p is prime. Not what you usually do, but it has an interesting solution.
oh hello there lol
Nice question. xD.
Since you could say that p is equal to a number 2ⁿ-2 and q is 1 therefore p^q+q^p is 2ⁿ-1. Since i know that it isnt determined whether there is an infinite amount of mersenne or not, answering this question, would be either quite impossible or just impossible.
A couple mersenne examples would be
(2,1): 3
(4,1) : 5
(6,1) : 7
.... You could generally say, that there are infinitely many tupels where p is a prime -1 and q is 1 so that (p -1)^1 + 1^(p-1) = p
So my answer would be that there are infinitely many tuples, as many as there are Prime numbers...
(2,3)...
wlog assume p=2 and eval mod3
Very easy ngl
@@zoomlogo lmao hi
**For those who want the tl;dr explanation:**
i^x = 2, so x = log base i of 2 = ln(2)/ln(i) by base-change. This is just ln(2)/(pi/2 i) = ln(4)/(pi * i).
Please don't say 90 degrees, as we are all adults now...I think I laughed a little more than I should have lol
😆
5:09 "Check this out" watch the word "note" at 144p, trippy.
this was such a fun video lol i love how happy you get
its a pleasure watching you. thanks
Great, that is fascinating 👍
Can you do a video on how to change the pens in your hand? Thanks for your wonderful videos
Just asking how did you become so good in maths? I saw your previous videos for doubts and you make questions easy.
Practice solving problems, read college textbooks, participate in competitions, watch and learn proofs. Over a few years you'll rack up so much intuition and knowledge if you practice right snd consistently.
Your videos are amazing, thanks professor!!! 🤗🤩🥳
Thank you, sir
you could also take the ilog of 2 and rewrite it as ln(2) / ln(i) = ln(2) / 0.5ln(-1) = 2ln(2) / pi i if im not mistaken
You are great teacher
Great job!
4:17 me too dawg glad we got one thing in common 💯
You are the best ❤
HI STEVE, CAN I GET A PROMO CODE FOR YOUR LAMBERT W FUNCTION PURPLE TSHIRT BECAUSE I WANT TO ORDER 100 OF IT AND THE SHIPPING PRICE IS 800 DOLLARS
Yes so true
i agree, the lambert w function shirt looks so good haha
bprp merch ftw
makes you feel smart
Could you send me an email blackpenredpen@gmail.com and let me know why you are ordering 100 t-shirts? I will see what I can do for you.
Very nice video wow I'm a really huge math fan and keep it up !
Thank you!
This is my humble request to whomsoever is reading, please consider my problem:::
By Euler's identity
=> e^iπ + 1 = 0
=> e^iπ = -1
Square both the sides
=> (e^iπ)² = (-1)²
=> e^2iπ = 1
take natural log of both the sides
=> ln(e^2iπ) = ln(1)
=> 2iπ = 0
Please explain😢😢😢
By the way I have a couple more such demonstrations that kinda contradicts the identity which I am unable to recall rn, although I also have worked with this a lot, it feels not a peaceful identity at somepoints, But I do remember that the problem in the eq comes right after squaring both the sides. I am not sure with all this as I did this a long time ago but whatever I wrote is what I remember rn, sorry if I wasted any time, please consider atleast replying 🙏🙏🙏
Here is what you wrote:
Square both the sides
=> (e^iπ)² = (-1)²
=> e^2iπ = 1
^^This is not true. For complex inputs, a^b^c is not always equal to a^(bc). The rule of complex analysis dealing with log and exponent branches says so. Or else, one can prove anything we want. Here's one of my personal favorites with this fallacy:
Suppose we have a real number a. Then, a = e^(ln a) by definition.
It follows that,
a = e^(1 * ln a)
= e^[(2iπ/2iπ)*(ln a)]
= e^[(2iπ)*(ln a/(2iπ))].
Applying our fallacy, we see that the above expression equals [e^(2iπ)]^[ln a/(2iπ)].
But e^(2iπ) = 1 by Euler's identity. Thus, we get:
a = 1^[ln(a)/(2iπ)]. But 1^x = 1, (unless you use the same fallacy!) so all a = 1. Thus, all real numbers are 1 (obviously not true).
BPRP actually did a video on this a while back, seeing if 1^x = 2 is possible. Long story short, that equation had no solutions. But, there was a solution to the equation 1 = 2^(1/x). Raising both sides to the power of (1/x) caused some issues with domain and range restrictions, so the solutions obtained were technically "extraneous."
@@ianzhou3998 Thank you very much for correcting me or rather teaching me the actual reason.
I just took a look at some articles, I didn't noticed the actual law which was a^b^c = a^(b×c) for these elements should belong to the Real world.
Well I was on the right path to find my mistake on my other such demonstrations that had the same mistake, when I said "the problem in the eq comes right after squaring both the sides".
Half knowledge is more dangerous than no Knowledge
Anyways thank you for explaining that much and telling about the video of 1^x=2 and sorry If I were to be silly. 😊
this kind of math is so interesting to me
I never took precalc or a calculus class
just college algebra
we only got a slight introduction to imaginary numbers so all of this baffles me
glad I don't need calculus for my degree 😅
I just discovered your channel and it IS GREAT!
Your enthusiasm is amazing, I had a math texher like that 25 years ago, you really remind him. (his name was Gustave😂)
I'm 38, love maths and I stopped at this level (high school math option in my country)
But because of life and the obstacles on the way, I never was able to pursue in polytechnic.
But I always kept a close link with mathematics and especially analysis. I litteraly do integrals during my free time, it's so beautiful..
My favorite is to trick arrogant people in the STEM fields with a "simple ∫ 1/(x^4 + 1) dx
Most people fall in the trap.
Anyway, I love your content and I'm gonna be watching a lot of it to stay sharp!
Thank you
Thank you very much!!
What trap?
@@DPME820 people think it's an Ln
lovely. Thank you. I will visit here whenever I got freetime like now.
Broo this is insaneee 😵
I shudder to think of the complexity of any maths problem that would require the use of a green pen in addition to the red, black and blue!😄
Stunning proof👍👍👍
Please make a video on how to solve any kind of ∑ problem...
I need to learn..
we can use ln(z)= ln(|z|) + i(2npi+theta) too
Our genius is back we amazing questions 😀
i^i. . . . .my favorite
Very kewl video....love the info
Is there a use to solving this for the more general form of e^i(pi/2+2npi) and show that you can produce an x that satisfies all infinitely many integer values of n?
I looked at that expression and rewrote the theta value as (pi+4npi)/2 to make it more comfortable, then I set (e^i(pi+4npi)/2)^x = 2 and followed the same process to isolate the x from the equation.
(e^i(pi+4npi)/2)^x = 2, n ∈ Z
x * i(pi+4npi)/2 = ln(2)
x = 2ln(2)/i(pi+4npi)
x = -2iln(2)/(pi+4npi)
Someone could ask "but what if we don't start with e^i(pi/2)? What if we start with e^i(5pi/2), or e^i(13pi/2)?" I asked that while watching, but I realized that other polar values of i can still be raised to a power that makes i^x = 2.
Edit: Thanks to another discussion I thought about possible problems here and the reason why the video's focus on the principal value was there, based on where ln(x) isn't well-defined. If anyone has input on where this does and doesn't work, that would be appreciated!
If you wanna make it work for non-positive complex numbers, just change the 4npi part.
The solution of i^x = -2 is 2n - 2iln(2)/π
Professor please make a video on tricks used to solve limits
Splendid.
I have known a lot about complex from your video ✨
The "secret" is always the same: put everything in base *e* (Euler's number)
I got the principal value for x just fine, but for the general solution I somehow ended up with 4n in the denominator.
Same :(
Can you make a video about the levi civita symbol, more specifically about some identities? Anyways, nice video, your work is appreciated!
I'm so prone to clickbait thumbnails when it comes to maths, usually they don't work on me but your thumbnails always get me😂
I took log base i on both sides and got x=logi(2)
Will you do hypercomplex numbers or quarternions?
Yes, it can. Take the natural log of both sides to get the exponent out, then divide by the ln i. X=ln 2 / ln i.
The term in the denominator is sort of kicking the problem back at you. 'e^x = i' is not at all clear to me.
I would love to see you go further than complex numbers and solve an equation with quaternions instead - maybe something like x^x = 2?
4:17 "i dont like to be on the bottom, i like to be on the top" xddd
Can you do 100 related rates please
Good one.
Good morning.
Is it possible for you to post your long videos in the video section of Facebook?
I would unload them from TH-cam but I have no credit card.
I would love to study each of your solutions now tht I have plenty of time for it (living at the foot of the northwestern argentinian mountains
At school
Teacher: Whats your favorite number?
A random kid: 3
Another kid: 7
This guy: *i*
@Blackpenredpen , look up this book called "
(Almost) impossible integrals, sums, and series" and do a video on it.
In case of the derivation of y= i^x to y´=x*i if x=2: i^2(Y)= 2(y´)
Bro you look so much better without a beard, no kidding
great video
I didn't get why we have to write i^4n ? thank you !
I just found this platform. So how does one start Calculus? Which video first?
0:07 😮
Before watching video
i^x = 2
x=log_{i}(2)
x=ln(2)/ln(i)
ln(i)=iπ(1+2n)/2 for any integer n
x=(2ln(2))/(iπ(1+2n))
x=(-2 i ln(2) )/( π (1 + 2n) )
focusing on the principle value, we have -2i ln(2) / π
Edit: shoot I didn't notice the extra answers with raising i to the fourth power
I did it this way too before watching the vid lol I also didn't notice the extra answers, also I think you made a slight mistake, ln(i) = iπ(1 + 4n)/2 after you combine the π/2 with the 2πn into a single fraction and factor out the π. It didn't affect the principal answer tho
Nice. Can you please also cover possible applications? I find it easier to remember that way!
One application is to exercise your math skills...
@@esajpsasipes2822 That's given right he's an effective communicator. I'm just asking for actual applications.
@@lucidx9443 I was more wanting to say that there might not be any (not sure), but even if there aren't any direct applications, it at least serves as an exercise (or entertainment i guess).
More generally, quantum mechanics uses this type of math
i^x=2 is the same as e^(½πix)=e^(ln2). So x=ln2⁄(½πi), or −i⋅ln4∕π
BPRP: x=4n-(2iln2)/π
My mindset: x=log_i(2)
-iLn(2)/pi
can u do vid solve equation
erf(x) = 2 find x?
Can u do a video where to use blackpen and redpen
i = e^(ipi/2), so (e^ipi/2)^x = e^x(ipi/2) = 2, take the ln
of both sides: we have x(ipi/2) = ln(2) => x = 2ln(2)/(ipi)
1:37 is i multiply by i allowed with index law(power to multiply) in complex world?
Request: is there a complex number x such that 2^x = x?
2^x = x
e^(x ln 2) = x
x e^(-x ln 2) = 1
-x ln 2 = W(-ln 2)
x = W(-ln 2) / -ln 2
Now there's a couple cool things of note here.
Any number such that n = W(-ln x) / -ln x can be represented as an infinite power tower. I'll post the proof in a separate reply underneath this one.
The other cool thing is that for 2^x = x, you can do 2^(2^x) = x or 2^2^x and following the chain, you're solving for the infinite power tower of twos
n^x = x
xln(n) = ln(x)
ln(n) = ln(x)e^(-ln(x))
W(-ln(n)) = -ln(x)
e^(-W(-ln(n))) = x
Identities of the W Lambert function tells us now that
x = W(-ln(n))/-ln(n)
For
n^x = x
So that's pretty cool, it's a shortcut to solve any convergent power towers.
Hey bprp, thank you for showing us all this crazy stuff, i am reeeeeeally enjoying while watching you
I learnt lambert w Function with you and i improved myself a lot, and for so long you didnt make a video about lambert W Function, i have little bit hard problem for you about this stuff; Can you solve x^(1/W(x))=y? I miss the w Function videos btw. I hope u see, thx again! (If you put in wolframalpha it may not show the solution but it is actually solveable, and the domains are suitable enough)
"We are adults now, so say 'pi over 2.'"
Thank you for this. From the bottom of my tired heart.
Sir, what's about taking log both sides...
Impressive
Où avez-vous trouver votre t-shirt ?
(4×ln2)/2πi is also a solution
brother i took both side to the power i then replaced i^i by e^-pi/2
then after something i got x = i^-4lni/pi
pls explain how much wrong i am
Hi Dr. Pen!
Cool!
4:16
you may have lost a part of solutions cause Ln2=ln2 + 2(pi)ki
In step two where you wrote (e^ipi/2)^x, isnt it incorrect to multiply the powers as one of them is complex? I saw it in another video where pi was falsely proven to be = 0 due to this mistake.
4:17 "I don't like to be on the bottom. I like to be on the top."
can u solve ln(4x+5)=4x-2016 ?
Can you have exact result for x, in x^i = i^x ?
Yes. x is either i or -i. Sorry if this is disappointing.
my man's hoarding whiteboard markers like they're Hagaromo chalk
😂
Could you please help me with a curiosity of mine?
Is there a way to create the exact formula for the following definite integral?
Integral between 0 and a:((((x^2)(b^2))/((a^4)(1-((x^2)/(a^2))))+1)^1/2) dx
Wolfram Alpha says that this is an indefinite integral - and the solution contains an Elliptic integral of second kind.
I wrote the answer as ln((2)^(2)(-i/pi)). I wonder if the whether general answer of i^x = a, would always be ln((a)^(a)(-i/pi)). It looks quite nice as well.
Try an induction proof
Tried it lol. Doesn't work. I did find after starting again that the actual general form is x = ln(a^(-2i/pi)) which I could prove by induction.
Thanks for the hint. @Simon N I don't get to do maths much these days as I have left sixth form and uni course doesn't have any advanced maths in it really alway fun when bprp uploads.
Why wolfpharm alfa give the answer with log not ln ?
yep solved
x=(2/i*2n+1*pie)ln2
n belongs to Integers.
i^x = 2
x = ln(2)/ln(i)
x = ln(2)/(i*(pi/2 + 2n*pi))
x = -iln(2)/(pi/2 + 2n*pi)
Where n is all integers
(arguably) more simple solution
I tought the same, but it seems this is not the same as what is in the video. I do not know how to pass form one to another, they should be the same
Consider a≠0≠b as complex numbers. Then a^x=b can be solved. Such is the power of the complex plane. And then, if one of a or b is equal to 0, the other must be as well in order to be solved.
Or a = 0 and b = 1, then x = 0, since 0^0 is (typically) defined as being 1.
@@simonwillover4175 yes, if that computational convention is followed, then that is the exception.
just take log_i of both sides
Hey, so i had an exam today, and 'this woman o_o' gave us a quiz online, everyone could get different thing to solve... I got to do an integral of x^100 * sinx dx... I've looked at it and was like "wtf...". Is there any 'clever' way to solve this or my only choice was doing integrating by parts / D I method (not all the way ofc o_o... I've made 5-6 multiplications and then + ... + C xD)
@blackpenredpen HELP (if it's even possible) XD
yeah, when x=log_i(2) :P
It is not clear to me that (x^y)^i = x^(iy). Clearly such is true if you look at an integer exponent, but it is NOT TRUE for general exponents. For a counter example consider f(t)=e^(i π t), which is clearly not identical to 1. However consider for x>0, f(x) = e^(i π 2t/2) = (e^(iπ))^2^(t/2) = ((-1)^2)^(t/2) = 1^(-t/2) = 1. This shows that sometimes it is false that x^(yz) = (x^y)^z.
It should be exp(z)^w=exp((z+2kπi)w), but general exponents a^z is defined by a^z=exp(z*ln(a)). In the video bprp is considering only the principal branch so it simplifies to exp(zw)=exp(z)^w.